Answer
$A=\frac{1}{2}$
Work Step by Step
Looking at the graph of the curve, we can see that the region goes from $0$ to $\pi/2$, so the bounds of the integral are $0$ and $\pi/2$.
Using the formula for the area under a polar curve, we can set up the integral:
$$\frac{1}{2}\int_{0}^{\pi/2} r^2 d\theta=\frac{1}{2}\int_{0}^{\pi/2} sin\,2\theta\, d\theta$$.
Solving the integral, we get:
$$\frac{1}{2}\bigg(-\frac{1}{2}cos\,2\theta\bigg)\bigg\rvert_{0}^{\pi/2}=\frac{1}{2}\Bigg(\frac{1}{2}-\bigg(-\frac{1}{2}\bigg)\Bigg)=\frac{1}{2}$$
